. The lines `(a+b)x + (a-b) y-2ab=0, (a-b)x+(a+b)y-2ab = 0 and x+y=0` form an isosceles triangle whose vertical angle is

The straight lines 3x+y-4=0,x+3y-4=0 and x+y-4=0 form a triangle which is :

If the straight lines 2x+3y-1=0,x+2y-1=0 ,and a x+b y-1=0 form a triangle with the origin as orthocentre, then (a , b) is given by (a) (6,4) (b) (-3,3) (c) (-8,8) (d) (0,7)

The condition on a and b , such that the portion of the line a x+b y-1=0 intercepted between the lines a x+y=0 and x+b y=0 subtends a right angle at the origin, is (a) a=b (b) a+b=0 (c) a=2b (d) 2a=b

If the straight lines 2x+3y-1=0,x+2y-1=0,and ax+by-1=0 form a triangle with the origin as orthocentre, then (a , b) is given by

Show that the straight line x^2-4xy+y^2=0 and x + y = 3 form an equilateral triangle.

In triangle ABC, the coordinates of the vertex A are , (4,-1) and lines x - y - 1 = 0 and 2x - y = 3 are the internal bisectors of angles B and C . Then the radius of the circles of triangle AbC is

x+y=7 and a x^2+2h x y+a y^2=0,(a!=0) , are three real distinct lines forming a triangle. Then the triangle is (a) isosceles (b) scalene (c) equilateral (d) right angled

Solve: (a - b )x + (a + b)y = a^(2) - 2ab - b^(2) and (a + b) (x + y) = a^(2) + b^(2)

Prove that (a b+x y)(a x+b y)>4a b x y(a , b ,x ,y >0)dot